This move was first explained
here.
IOE When the
Outies "see" all of a candidate for a house
except for the
Innies, the
3rd cell must
Equal the in/out difference (IOD).
i. For example, "45" n3: 1 outie r4c9 +
5 = 2 innies r3c78
ii. the 1 outie (r4c9) sees all of 4, 8 & 9 for n3 (they can't be in the 8(2)n3), except for the innies r3c78.
iii. so, when 8 is in r4c9, 8 must be in the innies, r3c78. Since 1 innie = 1 outie -> the 3rd cell (in this case, one of r3c78) must equal the IOD for n3 = 5
iv. same thing for 9 since it sees all 9s in n3 except r3c78. So, 9 in r4c9 -> r3c78 = {59}(IOE)
v. of course, this doesn't work for the 5 in r4c9 because it doesn't see all 5s in n3 apart from the innies. One 5 can hide from r4c9 in r1c8.
9. So, "45" n3: 1 outie r4c9 + 5 = 2 innies r3c78
9a. r4c9 = (589). If not 5, it must be 8 or 9 -> 5 locked in r3c78 (IOE).
9b. -> 5 locked in innies of n3 OR r4c9
9c. -> no 5 in common peers in r4c8, r1c7, r2c789, r3c9 & r1c9.
NOTE:
IOE is not enough on it's own to make eliminations in this case. Only if r4c9 = (89) would we have been able to lock 5 in r3c78 for n3 & r3 using
IOE only.
9e. no 8 in r4c9 (h13(2))
Thanks to
Afmob for suggesting putting this in the techniques forum
[edit: and to Andrew for some clarifications.].
Cheers
Ed